To factor trinomials like this one, we need to do a reverse FOIL. In other words, we need to find two binomials that multiply together to yield $\displaystyle 7x^{2} - 11x - 6$.

Finding the "first" terms is relatively easy; they need to multiply together to give us $\displaystyle 7x^{2}$, and since $\displaystyle 7$ only has two factors, we know the terms must be $\displaystyle 7x$ and $\displaystyle x$. We now have $\displaystyle (7x$    $\displaystyle )(x$    $\displaystyle )$, and this is where it gets tricky.

The second terms must multiply together to give us $\displaystyle -6$, and they must also multiply with the first terms to give us a total result of $\displaystyle -11x$. Many terms fit the first criterion. $\displaystyle (-2)(3)$, $\displaystyle (-3)(2)$, $\displaystyle (-6)(1)$ and $\displaystyle (-1)(6)$ all multiply to yield $\displaystyle -6$. But the only way to also get the "$\displaystyle x$" terms to sum to $\displaystyle -11x$ is to use $\displaystyle (7x + 3)(x - 2)$. It's just like a puzzle!

To find the greatest common factor, we must break each term into its prime factors:

$\displaystyle \small x^2y^3z^2 = x \cdot x \cdot y \cdot y \cdot y \cdot z \cdot z$

$\displaystyle \small x^4y^3z= x \cdot x \cdot x \cdot x \cdot y \cdot y \cdot y \cdot z$

The terms have $\displaystyle \small x^2$, $\displaystyle \small y^3$, and $\displaystyle \small z$ in common; thus, the GCF is $\displaystyle \small x^2y^3z$.

Pull this out of the expression to find the answer: $\displaystyle \small x^2y^3z(z+x^2)$. 

Use the $\displaystyle ac$-method to split the middle term into the sum of two terms whose coefficients have sum $\displaystyle -31$ and product $\displaystyle 6*35 = 210$. These two numbers can be found, using trial and error, to be $\displaystyle -21$ and $\displaystyle -10$.

$\displaystyle -21*-10=210$ and $\displaystyle -21+(-10)=-31$

Now we know that $\displaystyle 6x^{2}-31x+35$ is equal to $\displaystyle 6x^{2}-10x-21x+35$.

Factor by grouping.

$\displaystyle (6x^{2}-10x)-(21x-35)$

$\displaystyle 2x(3x-5)-7(3x-5)$

$\displaystyle (2x-7)(3x-5)$

First, we note that the coefficients have an LCD of 3, so we can factor that out:

$\displaystyle 3(x^{2} - 15x - 32)$

We try to factor further by factoring quadratic trinomial $\displaystyle x^{2} - 15x - 32$. We are looking to factor it into two factors$\displaystyle (x+ ? )(x+?)$, where the question marks are to be replaced by two integers whose product is $\displaystyle -32$ and whose sum is $\displaystyle -15$. 

We need to look at the factor pairs of $\displaystyle -32$ in which the negative number has the greater absolute value, and see which one has sum $\displaystyle -15$:

$\displaystyle \begin{matrix} \textrm{Pair} & \textrm{Sum} \\ 1,-32& -31\\ 2,-16&-14 \\ 4,-8& -4 \end{matrix}$

None of these pairs have the desired sum, so $\displaystyle x^{2} - 15x - 32$ is prime. $\displaystyle 3(x^{2} - 15x - 32)$ is the complete factorization.

When working with a rational expression, you want to first put your monomials in standard format.

Re-order the bottom expression, so it is now reads $\displaystyle -4x+2$. 

Then factor a $\displaystyle -1$ out of the expression, giving you $\displaystyle -1(4x-2)$.

 The new fraction is  $\displaystyle \frac{4x-2}{-1(4x-2)}$.

Divide out the like term, $\displaystyle (4x-2)$, leaving $\displaystyle \frac{1}{-1}$, or $\displaystyle -1$.

First pull out 3u from both terms.

3u⁴ – 24uv³ = 3u(u³ – 8v³) = 3u[u³ – (2v)³]

This is a difference of cubes. You will see this type of factoring if you get to the challenging questions on the GRE. They can be a pain to remember, but pat yourself on the back for getting to such hard questions! The difference of cubes formula is a³ – b³ = (a – b)(a² + ab + b²). In our problem, a = u and b = 2v:

3u⁴ – 24uv³ = 3u(u³ – 8v³) = 3u[u³ – (2v)³]

                = 3u(u – 2v)(u² + 2uv + 4v²)

This is a difference of squares. The difference of squares formula is a² – b² = (a + b)(a – b).

In this problem, a = 6x and b = 7y:

36x² – 49y² = (6x + 7y)(6x – 7y)

$\displaystyle 2x^{2}-7x+6$

$\displaystyle =2x^{2}-3x-4x+6$

$\displaystyle =2x^{2}-3x-(4x-6)$

$\displaystyle =x(2x-3)-2(2x-3)$

$\displaystyle =(x-2)(2x-3)$

$\displaystyle x^2 -10x - 24$

To factor, we are looking for two terms that multiply to give $\displaystyle -24$ and add together to get $\displaystyle -10$.

Possible factors of $\displaystyle -24$:

$\displaystyle (-1,24),\ (1,-24),\ (-2,12),\ (2,-12),\ (-3,8),\ (3,-8),\ (-4,6),\ (4,-6)$

Based on these options, it is clear our factors are $\displaystyle 2$ and $\displaystyle -12$.

$\displaystyle (2)(-12)=-24\ \text{and}\ 2+(-12)=-10$

Our final answer will be:

$\displaystyle (x-12)(x+2)$

$\displaystyle x^2 +7x - 8$

To factor, we are looking for two terms that multiply to give $\displaystyle -8$ and add together to get $\displaystyle 7$.

Possible factors of $\displaystyle -8$:

$\displaystyle (-1,8),\ (1,-8),\ (-2,4),\ (2,-4)$

Based on these options, it is clear our factors are $\displaystyle -1$ and $\displaystyle 8$.

$\displaystyle (-1)(8)=-8\ \text{and}\ (-1)+8=7$

Our final answer will be:

$\displaystyle (x+8)(x-1)$